Question

Container R holds a solution that is 20% alcohol and Container S holds a solution that is 60% alcohol. How many liters of solution from each container should be used to create 2 liters of solution that is 50% alcohol?

*make two equations*

Answer 1

The number of liters for container R = x = 0.5 liters

The number of liters for container S = y = 1.5 liters.

Explanation:

Let the number of liters of container R = x

Let the number of liters of container S = y

We are told in the question : that

Container R holds a solution that is 20% alcohol

= 20% × x = 0.20x

Container S holds a solution that is 60% alcohol.

= 60% × y = 0.60y

How many liters of solution from each container should be used to create 2 liters of solution that is 50% alcohol?

Hence;

x + y = 2 ....... Equation 1

x = 2 - y

0.20x + 0.60y = 2 × 50%

0.20x + 0.60y = 1........ Equation 2

Substitute

2-y for x in Equation 3

0.20(2 - y) + 0.60y = 1

0.40 - 0.20y + 0.60y = 1

Collect like terms

0.40y = 1 - 0.40

0.40y = 0.60

y = 0.60/0.40

y = 1.5 liters

Note that :

x = 2 - y

x = 2 - 1.5

x = 0.5 liters

Therefore,

the number of liters for container R = x = 0.5 liters

the number of liters for container S = y = 1.5 liters.